clc408 ETC-unknow, clc408 Datasheet - Page 7

• Manufacturer Part Number: clc408
• Description: High-speed, Low-power Line Driver
• Manufacturer: ETC-unknow

For non-inverting and transimpedance gain configurations:
For inverting gain configurations:
Capacitive Loads
Capacitive loads, such as found in A/D converters,
require a series resistor (R
settling performance. The Settling Time vs. Capacitive
Load plot in the Typical Performance Characteristics
section provides the information for selecting this resistor.
Using a resistor in series with a reactive load will also
reduce the load’s effect on amplifier loop dynamics.
For instance, driving coaxial cables without an output
series resistor may cause peaking or oscillation.
Transmission Line Matching
One method for matching the characteristic impedance
of a transmission line is to place the appropriate
resistor at the input or output of the amplifier. Figure 6
shows the typical circuit configurations for matching
transmission lines.
In non-inverting gain applications, R
directly to ground. The resistors R
V
V
1
2
+
+
-
-
R
R
Figure 6: Transmission Line Matching
1
4
Capacitance across R
A capacitive load
Long traces and/or lead lengths between R
and the CLC408
Extra capacitance between the inverting
pin and ground (C
Inadequate ground plane at the non-inverting
pin and/or long traces between non-inverting
pin and ground
capacitance for R
and a capacitive load (see the Settling
Time vs. C
sub-section below for suggestions on
reducing C
after reducing C
non-inverting pin and ground (see R
Figure 2)
Do not place a capacitor across R
Use a resistor with low parasitic
Use a series resistor between the output
Keep these traces as short as possible
See the Printed Circuit Board Layout
Increase R
Place a 50 to 200
Z
Z
0
0
R
R
2
5
R
R
3
g
g
f
L
if peaking is still observed
g
plot)
)
+
CLC408
-
s
) in the output to improve
g
R
f
f
f
resistor between the
1
, R
C
R
6
6
2
g
, R
is connected
6
, and R
Z
0
t
in
f
7
f
are
R
7
V
o
7
equal to the characteristic impedance, Z
mission line or cable. Use R
from reactive loading caused by the transmission line,
or by parasitics.
In inverting gain applications, R
ground. The resistors R
The parallel combination of R
The input and output matching resistors attenuate the
signal by a factor of 2, therefore additional gain is needed.
Use C
frequency range. It compensates for the increase of
the op amps output impedance with frequency.
Thermal Design
To calculate the power dissipation for the CLC408,
follow these steps:
To calculate the maximum allowable ambient tempera-
ture, solve the following equation: T
to ambient in °C/W, and T
Thermal Resistance section contains the thermal
resistance for various packages.
Dynamic Range (input /output protection)
ESD diodes are present on all connected pins for pro-
tection from static voltage damage. For a signal that
may exceed the supply voltages, we recommend using
diode clamps at the amplifier’s input to limit the signals
to less than the supply voltages.
The CLC408’s output current can exceed the maximum
safe output current. To limit the output current to <
96mA:
V
the maximum safe output current.
Dynamic Range (input /output levels)
The Electrical Characteristics section specifies the
Common-Mode Input Range and Output Voltage
Range; these voltage ranges scale with the supplies.
Output Current is also specified in the Electrical
Characteristics section.
JA
o(max)
, where
1) Calculate the no-load op amp power:
2) Calculate the output stage’s RMS power:
3) Calculate the total op amp RMS power:
6
to match the output transmission line over a greater
Limit the output voltage swing with diode
clamps at the input
Make sure that
P
P
P
is the output voltage swing limit, and I
amp
o
t
= P
= (V
are the RMS voltage and current across the
external load
JA
= I
amp
CC
is the thermal resistance from junction
CC
+ P
– V
•
(V
o
load
CC
4
R
, R
)
L
– V
amb
•
5
I
6
load
, and R
and R
EE
3
V
3
I
is in °C. The Package
o(max)
o(max)
to isolate the amplifier
is connected directly to
)
, where V
g
is also equal to Z
7
amb
http://www.national.com
are equal to Z
o
, of the trans-
load
= 175 – P
and I
o(max)
load
t
o
o
is
.
.
•